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• 评论了博文 Golang Once源码解析
  先将done设置为1，再执行f函数 这个说反了吧？ // Do guarantees that when it returns, f has finished. // This implementation would not implement that guarantee: // given two simultaneous calls, the winner of the cas would // call f, and the second would return immediately, without // waiting for the first's call to f to complete. // This is why the slow path falls back to a mutex, and why // the atomic.StoreUint32 must be delayed until after f returns.